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Proof step 2 from proof bits2int-rec-01 (previous step)

bits.first ⊦ bits = [], ~1 = bits2uint(bitsinvert(bits.rest)) * 2 + bits2int(bits + true ')

Heuristic var elimination applied rule insert elim lemma

And closed the goal!

Rule argument

seq: bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0 subst: [bits, boolvar, bits0] → [bits, boolvar, bits0] rotate: T spec: list-data inst: bitlista name: elim-list-c

Additional information

Used simplifier rules: bits2int-neg: ⊦ bits2int(true ' + bits) = ~ bits2uint(bitsinvert(true + bits) +1);%bits2uint-zeros-03: ⊦ bits2uint(false ' + bits) = bits2uint(bits);%bits2uint-one: ⊦ bits2uint(bits + true ') = bits2uint(bits) * 2 + 1;% ⊦ i * j = j * i;% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ boolvar ' + bits ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bitsinvert(true ' + bits) = false + bitsinvert(bits);% ⊦ bitsinvert(bits + bits0) = bitsinvert(bits) + bitsinvert(bits0);% ⊦ bitsinvert(true ') = false ';% ⊦ (bits + false ')+1 = bits + true;% ⊦ bits + boolvar = bits + boolvar ';% ⊦ i + ~ j = i - j;% ⊦ i - j + i = ~ j;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;%bits ≠ [] ⊦ (boolvar ↔ bits.first) ∧ bits0 = bits.rest ↔ bits = boolvar + bits0; (elim) Used terms: bits, boolvar, bits0